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Mathematics 18 Online
OpenStudy (anonymous):

evaluate: ∫(9/(x^2+9))dx the answer is 3arctan(x/3) + 3...how? how do you get the 3 and x/3 ?

OpenStudy (mimi_x3):

\[\int\limits\frac{9}{x^{2}+9} dx =>9\int\limits\frac{1}{x^{2}+3^{2}} dx\] Inverse trig function..

OpenStudy (mimi_x3):

\[\int\limits\frac{1}{a^{2}+x^{2}} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} \]

OpenStudy (apoorvk):

Mimi knows so much about integration at the age of 10? \[\huge \color{red}{\normalsize\text{P}}\color{orange}{\normalsize\text{R}}\color{#9c9a2e}{\normalsize\text{O}}\color{green}{\normalsize\text{D}}\color{blue}{\normalsize\text{I}}\color{purple}{\normalsize\text{G}}\color{purple}{\normalsize\text{Y}}\color{red}{\normalsize\text{ }}\color{orange}{\normalsize\text{I}}\color{#9c9a2e}{\normalsize\text{ }}\color{green}{\normalsize\text{T}}\color{blue}{\normalsize\text{O}}\color{purple}{\normalsize\text{L}}\color{purple}{\normalsize\text{D}}\color{red}{\normalsize\text{ }}\color{orange}{\normalsize\text{Y}}\color{#9c9a2e}{\normalsize\text{O}}\color{green}{\normalsize\text{U}}\color{blue}{\normalsize\text{^}}\color{purple}{\normalsize\text{^}}\color{purple}{\normalsize\text{!}}\color{red}{\normalsize\text{}}\]

OpenStudy (mimi_x3):

lol, i thought that you were going to elaborate the answer. xD Me no prodigy. :P

OpenStudy (anonymous):

oh...sank u very much. I have been enlightened and now i can move on :) (i was wondering how my textbook just expected us to know that...)

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